Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long a, b, c, d, i, e, f, g, n, m, k, l, in[200005], out[200005],
    C[200005], S[200005], B[22][200005], ver, u, u1, q;
vector<long long> v[200005];
struct node {
  int A[12];
  node() { memset(A, 63, sizeof(A)); }
  void insert(long long val) {
    A[11] = val;
    sort(A + 1, A + 12);
  }
} vals[22][200005];
node merge(node x, node y) {
  node ans = x;
  for (long long i = 1; i <= 11; i++) ans.insert(y.A[i]);
  return ans;
}
void dfs(long long x, long long y, long long z) {
  k++;
  in[x] = k;
  C[x] = z;
  B[0][x] = y;
  for (long long i = 1; i <= 20; i++) {
    B[i][x] = B[i - 1][B[i - 1][x]];
    vals[i][x] = merge(vals[i - 1][x], vals[i - 1][B[i - 1][x]]);
  }
  for (long long i = 0; i < v[x].size(); i++) {
    if (v[x][i] == y) continue;
    dfs(v[x][i], x, z + 1);
  }
  out[x] = k;
}
long long LCA(long long x, long long y) {
  if (in[x] <= in[y] && out[y] <= out[x]) return x;
  for (long long j = 20; j >= 0; j--) {
    if ((in[B[j][x]] <= in[y] && out[y] <= out[B[j][x]]) ||
        (in[B[j][x]] == 0 && out[B[j][x]] == 0)) {
      continue;
    } else {
      x = B[j][x];
    }
  }
  return B[0][x];
}
node go(long long v, long long x) {
  node ans;
  for (long long i = 0; i <= 20; i++)
    if ((1 << i) & x) {
      ans = merge(ans, vals[i][v]);
      v = B[i][v];
    }
  return ans;
}
int main() {
  cin >> n >> m >> q;
  for (long long i = 1; i < n; i++) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (long long i = 1; i <= m; i++) {
    cin >> a;
    vals[0][a].insert(i);
  }
  dfs(1, 0, 1);
  for (long long i = 1; i <= q; i++) {
    cin >> a >> b >> c;
    k = LCA(a, b);
    node x = go(a, C[a] - C[k]);
    node y = go(b, C[b] - C[k] + 1);
    node ans = merge(x, y);
    l = 0;
    while (l < c && ans.A[l + 1] <= m) {
      l++;
    }
    c = l;
    cout << c << " ";
    for (long long i = 1; i <= c; i++) cout << ans.A[i] << " ";
    cout << endl;
  }
}